Dividing with a square root: 2x^3 + sqrt3x^3 + 15x + ....

[Guest]

My homework equation says to divide

2x^3 + (sqrt3x^2) + 15x + (6 times sqrt3) = 0

Given = (sqrt3) as a root.

How in the world do I even start this problem?

The answers to this problem include.

A. sqrt3, negative sqrt3, -2
B. sqrt3, -2(sqrt3), 2(sqrt3)
C. sqrt3, (sqrt3) / 2, (negative sqrt3) / 2
D. sqrt3, -2(sqrt3), (sqrt3) / 2

 

[stapel]

Are you really supposed to "divide"? If so, by what?

Or do the instructions say something more along the lines of "Find all the solutions, given that x = sqrt[3] is a root, by using synthetic division"...?

Also, is the equation any of the following?

. . . . .2x³ + sqrt[3x²] + 15x + 6sqrt[3] = 0

. . . . .2x³ + sqrt[3x]² + 15x + 6sqrt[3] = 0

. . . . .2x³ + sqrt[3]x² + 15x + 6sqrt[3] = 0

Or is it something else?

Thank you.

Eliz.

 

[Guest]

Whoops!

The equation is:

. . . . .2x³ + sqrt[3x²] - 15x + 6sqrt[3] = 0

And the instructions say to solve. Sorry for not making that clear the first time and it is minus 15x not plus 15x like I put in the first post.

 

[Denis]

. . . . .2x³ + sqrt[3x²] - 15x + 6sqrt[3] = 0

sqrt(3x^2) ? Can't be; should be sqrt(3)x^2; RE-CHECK :shock:

 

[stapel]

The equation is: 2x³ + sqrt[3x²] - 15x + 6sqrt[3] = 0

Since sqrt[x²] = |x|, the equation reduces to (the somewhat odd):

. . . . .2x³ + sqrt[3] |x| - 15x + 6sqrt[3] = 0

When dealing with absolute values, it is often best to consider cases. So, letting x < 0, we have:

. . . . .2x³ - sqrt[3]x - 15x + 6sqrt[3] = 0

. . . . .2x³ - (sqrt[3] + 15)x + 6sqrt[3] = 0

Unfortunately, I don't see any simple way of solving this, and we can't use the provided root, since it doesn't apply to this half of the function.

Turning to the other case, we have x > 0, which gives:

. . . . .2x³ + sqrt[3]x - 15x + 6sqrt[3] = 0

. . . . .2x³ + (sqrt[3] - 15)x + 6sqrt[3] = 0

We are given that x = sqrt[3] is one solution. Once we divide this out (by synthetic division or whatever other method), we will be left with a quadratic, to which we can apply the Quadratic Formula.

However, this solution does not appear to be valid:

. . .2[sqrt[3]]³ + (sqrt[3] - 15)[sqrt[3]] + 6sqrt[3]

. . . . .= 2(3)sqrt[3] + 3 - 15sqrt[3] + 6sqrt[3]

. . . . .= 6sqrt[3] + 3 - 15sqrt[3] + 6sqrt[3]

. . . . .= 12sqrt[3] - 15sqrt[3] + 3

. . . . .= 3 - 3sqrt[3]

Since this does not equal zero, we are still stuck. Sorry.

Eliz.

 

[skeeter]

\(\displaystyle \L 2x^3 + \sqrt{3}x^2 - 15x + 6\sqrt{3} = 0\)

\(\displaystyle \sqrt{3}\) is a root ... use synthetic division

sqrt(3)].............2.............sqrt(3).............. -15 ............6sqrt(3)
..................................2sqrt(3).................9............-6sqrt(3)
----------------------------------------------------------------------------
.....................2............3sqrt(3)................-6...............0

\(\displaystyle \L 2x^3 + \sqrt{3}x^2 - 15x + 6\sqrt{3} = (x - \sqrt{3})(2x^2 + 3\sqrt{3}x - 6) = 0\)

use the quadratic formula to find the two roots of the quadratic factor.

 

[Guest]

\(\displaystyle \L 2x^3 + \sqrt{3}x^2 - 15x + 6\sqrt{3} = 0\)

\(\displaystyle \sqrt{3}\) is a root ... use synthetic division

sqrt(3)].............2.............sqrt(3).............. -15 ............6sqrt(3)
..................................2sqrt(3).................9............-6sqrt(3)
----------------------------------------------------------------------------
.....................2............3sqrt(3)................-6...............0

\(\displaystyle \L 2x^3 + \sqrt{3}x^2 - 15x + 6\sqrt{3} = (x - \sqrt{3})(2x^2 + 3\sqrt{3}x - 6) = 0\)

use the quadratic formula to find the two roots of the quadratic factor.

\(\displaystyle (x - \sqrt{3})(2x^2 + 3\sqrt{3}x - 6) = 0\)

Alright. I did get this far earlier, but I thought I had it completely wrong.

Which equation do I use to plug into the quadratic formula?

I would assume it's the second one, but then where does the \(\displaystyle x-\sqrt{3}\) come in?


:-l still confused!

 

[stapel]

To learn what the Quadratic Formula is and how to use it (something which should have been covered well before this point in your studies), try using a search engine. There are many very good lessons available.

The "x - sqrt[3]" came from the given: that x = sqrt[3] was a root. To learn how roots and factors relate, try studying the "Factor Theorem", or "how to find polynomials from their roots.

By the way, if you aren't familiar with synthetic division and how roots relate to factors, by what method did you "get this far [the factored form] earlier"? (We'll be glad to try to explain things using whatever method you are using, but you will need to tell us what that method is.)

Thank you.

Eliz.

 

[Denis]

Here's a simpler problem like the one you have:

x^3 + 2x^2 - 5x - 6 = 0
(x + 3)(x + 1)(x - 2) = 0 : x = -3 or -1 or 2

So the equation x^3 + 2x^2 - 5x - 6 = 0 is the result of:
(x + 3) times (x + 1) times (x - 2)

As example, if you divide (x^3 + 2x^2 - 5x - 6) by (x + 3),
result is x^2 - x - 2

If you solve x^2 - x - 2 = 0, you get (x + 1)(x - 2) = 0 : x = -1 or 2

Your problem:
"My homework equation says to divide
2x^3 + (sqrt3x^2) + 15x + (6 times sqrt3) = 0
Given = (sqrt3) as a root. "

If the example I used had been the one, then your problem would be:
My homework equation says to divide
x^3 + 2x^2 - 5x - 6 = 0
Given = -3 as a root.

Do you follow all that?
So your problem is really "as easy" as my example...
except the use of sqrt(3) makes it APPEAR complicated.